Are there any published studies on cases of infinite sums for which the Euler–Maclaurin summation method yields the exact evaluation?

Question

The Euler–Maclaurin summation formula is as follows: $$\sum_{i=m}^{n} f(i) = \int_{m}^{n} f(x) dx + \frac{f(n)+f(m)}{2} + \sum_{k=1}^{\lfloor p/2 \rfloor} \frac{B_{2k}}{(2k)!}\big{(}f^{(2k-1)}(n)-f^{(2k-1)}(m)\big{)} + R_{p} .$$

Usually, it yields an approximation, because the third term (the sum involving the Bernoulli numbers) and the remainder term cannot be evaluated exactly.

If the higher derivatives eventually become zero at the start and end points, the formula becomes exact. An example of an application of this fact is Faulhaber's formula.

However, Faulhaber's formula pertains to finite sums. I wonder whether there are infinite series for which all terms of the formula above can be evaluated. In particular, I would like to know whether some systematic study on the circumstances under which such an evaluation is possible and exact has been published.

I've asked a similar question In what case(s) does the Euler-Maclaurin summation method yield the exact evaluation? a while ago on MSE.

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Added note. To clarify, by ‘exact’ I mean ‘has a closed form expression’.

Answer 1

I am not sure this answer your question, but it is an approximation. It is not a reference, only that
many years ago I wrote for my use an exposition of Euler-MacLaurin's formula. Defining first the Bernoulli polynomials so that I can use them in successive integrations by parts. I arrived to the expression (with $a$ and $b$ integers) \begin{multline*} \sum_{j=a}^bf(j)=\int_a^bf(x)\,dx+\frac{f(a)+f(b)}{2}\\+\sum_{j=2}^n(-1)^j\frac{B_j}{j!}\bigl(f^{(j-1)}(b)-f^{(j-1)}(a)\bigr)+\frac{(-1)^{n+1}}{n!}\int_a^bf^{(n)}(x)\widetilde{B}_n(x)\,dx. \end{multline*} To determine the Bernoulli numbers, I proposed to apply the above to $f(x)=e^{-hx}$ with $h>0$; let $a=0$ and $b=m$. Then $f^{(j)}(x)=(-h)^je^{-hx}$, The right hand side is a geometrical series with sum $$\sum_{j=0}^m e^{-hj} =\frac{1-e^{-(m+1)h}}{1-e^{-h}}.$$ Hence our equality is in this case \begin{multline*} \frac{1-e^{-(m+1)h}}{1-e^{-h}}=\frac{1-e^{-mh}}{h}+\frac{1+e^{-mh}}{2}\\+ \sum_{j=2}^n(-1)^j\frac{B_j}{j!}\bigl[(-h)^{j-1}e^{-mh}-(-1)^{j-1}\bigr] +\frac{(-1)^{n+1}}{n!}\int_0^m(-h)^ne^{-hx}\widetilde{B}_n(x)\,dx. \end{multline*} Multiplying by $h$ and simplifying yields $$\frac{h}{1-e^{-h}}(1-e^{-(m+1)h})=\Bigl(\sum_{j=0}^n\frac{B_j}{j!}h^j\Bigr)(1-e^{-mh})-\frac{h^{n+1}}{n!}\int_0^m e^{-hx}\widetilde{B}_n(x).$$ Since $h>0$ we may take limits when $m\to+\infty$ and we get $$\frac{h}{1-e^{-h}}=\sum_{j=0}^n\frac{B_j}{j!}h^j-\frac{h^{n+1}}{n!}\int_0^m e^{-hx}\widetilde{B}_n(x).$$ So that for each $n$ there is a constant $C_n$ such that for $0<h<1$ $$\Bigl|\frac{h}{1-e^{-h}}-\sum_{j=0}^n\frac{B_j}{j!}h^j\Bigr|\le C_nh^{n+1}.$$ It follows that $x/(1-e^{-x})$ have a Taylor series $$\frac{x}{e^x-1}=\sum_{j=0}^\infty \frac{B_j}{j!}x^j.$$ This can be thought as an exact MacLaurin evaluation with $n=\infty$.